Assume we have a Hida test function $\varphi\in (\mathcal S)$, and $y\in \mathcal S'(\mathbb R)$. Define the Gateaux directional derivative of $\varphi$ (in the direction of $y$) by: $$D_y\varphi(x):=\sum_{n=1}^{\infty} n \langle :x ^{\otimes (n-1)}:,\langle y, f_n\rangle\rangle,$$ where the sequence of (symmetric) $f_n$'s is the element in the Fock space that corresponds to $\varphi$ (via the Ito-Segal isomorphism). At this point I need to calculate the $S$-transform of this element, but I am having some problems (probably trivial stuff) with the calculations: I have that by definition of the $S$-transform we have $$S(D_y\varphi(x))(\xi)=\sum_{n=1}^{\infty} n\big\langle \langle y,f_n\rangle, \xi^{\otimes (n-1)}\big\rangle.$$ From this I should be able to obtain $$\sum_{n=1}^{\infty} n\big\langle y\hat{\otimes} \xi^{\otimes (n-1)},f_n \big\rangle,$$ but this very last step is not clear to me and I haven't been able to find an explanation. N.Obata justifies this step by using contraction tensors but prefer to avoid that formulation. Could you please explain me this? I have the feeling that it's something straightforward but I haven't been able to get it. Thanks in advance! EDIT: A colleague told me that this could be a consequence of the Kernel Theorem , but honestly I don't see how. EDIT 2: So this is very informal but if one thinks of the pairing $\langle \cdot,\cdot\rangle$ as the inner product in $(L^2)$ (which is not entirely correct since one of the element could be outside $(L^2)$) we can write that: \begin{align} &\big\langle \langle y,f_n\rangle, \xi^{\otimes (n-1)}\big\rangle\\ =&\int_{S'(\mathbb R^{n-1})} \int_{S'(\mathbb R)} f_n(t_1,\cdots,t_n)y(t_1)\xi^{\otimes (n-1)}(t_2,\cdots,t_n) d\mu(t_1)d\mu^{(n-1)}(t_2,\cdots,t_n)\\ =&\int_{S'(\mathbb R^{n})}f_n(t_1,\cdots,t_n)y(t_1)\xi^{\otimes (n-1)}(t_2,\cdots,t_n) d\mu^n(t_1,t_2,\cdots,t_n)\\ =&\langle y\otimes \xi^{\otimes (n-1)},f_n\rangle, \end{align} But again this is informal and I don't know if I can justify all steps, what do you think?